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Alexxx [7]
3 years ago
15

HELP ASAP!! WILL TRY TO GIVE BRAINLIEST Describe chromatin and how it changes.

Physics
2 answers:
larisa86 [58]3 years ago
5 0

Answer: Me again! Chromatin is a complex of DNA and proteins that forms chromosomes within the nucleus of eukaryotic cells. Nuclear DNA does not appear in free linear strands; it is highly condensed and wrapped around nuclear proteins in order to fit inside the nucleus. Chromatin exists in two forms.

Explanation: Hope this helps! Happy Holidays!

White raven [17]3 years ago
3 0

Answer:

Chromatin is a mass of genetic material composed of DNA and proteins that condense to form chromosomes during eukaryotic cell division. Chromatin is located in the nucleus of our cells . The primary function of chromatin is to compress the DNA into a compact unit that will be less voluminous and can fit within the nucleus.

I hope this helps

Explanation:

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If a car travels at 45.0 m/s for 2 minutes. What is the car's distance?
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the answer is 90 m/s

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What is the definition of evolution
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How does the radius of a string affect centripetal force.
KiRa [710]

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because a raduis is half of 25% of a cicrle.

Explanation:

5 0
3 years ago
Lonnie pitches a baseball of mass 0.500 kg. The ball arrives at home plate with a speed of 35.0 m/s and is batted straight back
Andreyy89

Answer:

Explanation:

The impulse equation is

Δp = FΔt, where Δp = final momentum - initial momentum, F is the Force exerted on an object, and Δt is the change in time. In this equation,the entire right side defines the impulse. In other words, FΔt is the impulse; thus the change in momentum an object experiences is due to its change in impulse and is directly proportional to it.

Therefore, once we find the change in momentum, that is the impulse the object experiences. Δp = final momentum - initial momentum, where

p = mv and p is momentum.

p_f=(.500)(50.0) so

p_f=25.0 and

p_i=(.500)(35.0) so

p_i=17.5; therefore,

Δp = 25.0 - 17.5 = 7.5\frac{kg*m}{s} which is the unit for momentum

7 0
3 years ago
A 910-kg object is released from rest at an altitude of 1200 km above the north pole of the Earth. If we ignore atmospheric fric
gayaneshka [121]

In order to develop this problem it is necessary to use the concepts related to the conservation of both potential cinematic as gravitational energy,

KE = \fract{1}{2}mv^2

PE = GMm(\frac{1}{r_1}-(\frac{1}{r_2}))

Where,

M = Mass of Earth

m = Mass of Object

v = Velocity

r = Radius

G = Gravitational universal constant

Our values are given as,

m = 910 Kg

r_1 = 1200 + 6371 km = 7571km

r_2 = 6371 km,

Replacing we have,

\frac{1}{2} mv^2 =  -GMm(\frac{1}{r_1}-\frac{1}{r_2})

v^2 =  -2GM(\frac{1}{r_1}-\frac{1}{r_2})

v^2 = -2*(6.673 *10^-11)(5.98 *10^24) (\frac{1}{(7.571 *10^6)} -\frac{1}{(6.371 *10^6)})

v = 4456 m/s

Therefore the speed of the object when striking the surface of earth is 4456 m/s

3 0
4 years ago
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